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Contents · 0% complete
  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
LearnYMA01WME03 · Mechanics 3Further dynamicsVariable forces and differential equations
Further dynamics · Subtopic 1 of 1

Motion Under a Variable Force

When force changes, Newton's law becomes a differential equation. Write a=dvdta=\frac{dv}{dt} if you want time, or a=vdvdxa=v\frac{dv}{dx} if you want distance, separate the variables and integrate with the initial conditions.

Cambridge 9231 Paper 3Edexcel YMA01 WME03Edexcel 9FM0 Option≈ 50 minRevision 1 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • Choose dvdt\frac{dv}{dt} or vdvdxv\frac{dv}{dx} for acceleration.
  • Set up and solve separable equations for F(t)F(t), F(v)F(v), F(x)F(x).
  • Solve problems with resistance proportional to vv or v2v^2.
Pause and think

A boat cuts its engine. Water resistance is proportional to speed. Does it ever stop?

Section 2 of 14

Core idea

Start hereCoasting to a stop

A motorboat cuts its engine. Water drag is proportional to speed, so as the boat slows, the drag weakens too. Does the boat ever stop? Newton's second law, written as a differential equation, gives a surprising answer.

The big idea

With variable forces, F=maF=ma is a differential equation. Use a=dvdta=\frac{dv}{dt} to connect vv and tt, or a=vdvdxa=v\frac{dv}{dx} to connect vv and xx. Separate, integrate, apply initial conditions.

1. Two forms of acceleration

Where vdvdxv\frac{dv}{dx} comes from
  1. a=dvdta=\frac{dv}{dt}
    Definition.
  2. =dvdx⋅dxdt=\frac{dv}{dx}\cdot\frac{dx}{dt}
    Chain rule.
  3. =vdvdx=v\frac{dv}{dx}
    dxdt=v\frac{dx}{dt}=v.
Use this whenever the force depends on xx, or the question asks about distance.
Worked exampleResistance proportional to v

A 22 kg particle has only resistance 4v4v; v=10v=10 at t=0t=0. Find v(t)v(t).

Goal
vv in terms of tt.
Strategy
dvdt\frac{dv}{dt} form, separate.
  1. 2dvdt=−4v2\frac{dv}{dt}=-4v
    Newton's second law; resistance opposes motion.
  2. ∫dvv=∫−2 dt\int\frac{dv}v=\int-2\,dt
    Separate.
  3. ln⁡v=−2t+c, c=ln⁡10\ln v=-2t+c,\ c=\ln10
    Integrate; use v(0)=10v(0)=10.
Answerv=10e−2tv=10e^{-2t}.
✓ CheckNever reaches 0; total distance ∫0∞10e−2tdt=5\int_0^\infty10e^{-2t}dt=5 m, finite.

2. Force depending on position

Worked exampleInverse square repulsion

A 11 kg particle is repelled from OO by 8x2\frac8{x^2} N. At rest at x=1x=1. Find vv at x=4x=4.

Goal
vv.
Strategy
vdvdxv\frac{dv}{dx} form.
  1. vdvdx=8x2v\frac{dv}{dx}=\frac8{x^2}
    F=maF=ma, m=1m=1.
  2. 12v2=−8x+c\tfrac12v^2=-\frac8x+c
    Integrate both sides.
  3. 0=−8+c⇒c=80=-8+c\Rightarrow c=8
    v=0v=0 at x=1x=1.
  4. 12v2=−2+8=6\tfrac12v^2=-2+8=6
    At x=4x=4.
Answerv=12≈3.46v=\sqrt{12}\approx3.46 m s−1^{-1}.
✓ CheckWork done ∫148x−2dx=6\int_1^48x^{-2}dx=6 J == KE gained ✓.
Choosing the form

Let the question decide the form.

Common misconceptionSigns Resistance acts against motion: −kv-kv or −kv2-kv^2 when v>0v>0 in the positive direction.
Key takeawayTerminal velocity Falling with resistance kv2kv^2: acceleration is zero when kv2=mgkv^2=mg, giving vT=mgkv_T=\sqrt{\frac{mg}k}.
Pause and think

With resistance kv2kv^2 (no other force), does the particle travel a finite distance?

Summary

  • a=dvdt=vdvdxa=\frac{dv}{dt}=v\frac{dv}{dx}.
  • Separate and integrate.
  • Initial conditions fix constants.
Key words
Variable force
Force that changes with time, speed or position.
Terminal velocity
Constant speed when forces balance.
HistoryNewton again. Book II of Newton's Principia (1687) is devoted to motion in resisting media, with resistance proportional to vv and to v2v^2.
Fun factTerminal velocity. A skydiver reaches about 5555 m s−1^{-1} belly-down, when air resistance kv2kv^2 equals weight.
Section 3 of 14

Key formulae

Accelerationa=dvdt=vdvdx=d2xdt2a=\frac{dv}{dt}=v\frac{dv}{dx}=\frac{d^2x}{dt^2}
Newtonmdvdt=F(t,v)ormvdvdx=F(x,v)m\frac{dv}{dt}=F(t,v)\quad\text{or}\quad mv\frac{dv}{dx}=F(x,v)
TrickWhich form? Question mentions time → dvdt\frac{dv}{dt}. Mentions distance → vdvdxv\frac{dv}{dx}. Force depends on xx → always vdvdxv\frac{dv}{dx}.
Section 4 of 14

Worked examples

Example 1 · Resistance kv0 / 3 steps
A 22 kg particle has resistance 4v4v and no other force; v=10v=10 at t=0t=0. Find vv at time tt.
    Example 2 · Force of x0 / 3 steps
    A 11 kg particle feels force 8x2\frac{8}{x^2} away from OO. At x=1x=1, v=0v=0. Find vv at x=4x=4.
      Section 5 of 14

      Explore the graph

      Speed under resistance kv
      Loading the interactive graph…
      Try this
      1. xx is time. The curve never reaches zero.
      2. Increase mm: heavier bodies slow down more gradually.

      v=ue−kt/mv=ue^{-kt/m}.

      In the real worldRocket launches. Rocket thrust is roughly constant, but mass falls as fuel burns and drag depends on speed and height: engineers solve these equations numerically.
      Section 6 of 14

      Common mistakes

      IncorrectUsing suvat with variable force
      CorrectSolve the differential equation
      suvat needs constant acceleration.
      IncorrectResistance with the wrong sign
      CorrectOpposes motion: −kv-kv
      It acts against the velocity.
      IncorrectForgetting the constant
      CorrectUse initial conditions
      Each integration needs one.

      Force depends on x

      v dv/dx enters the chat.

      Section 7 of 14

      Exam tip

      Exam tipSeparate early. Get all vv terms with dvdv and all tt or xx terms with dtdt or dxdx before integrating.
      Section 8 of 14

      Guided practice

      SA-FM-VF-G013 marksL2 StandardNon-calculatorF(t), F(v)
      A 22 kg particle starts from rest under force F=6tF=6t N. Find its speed at t=2t=2.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-FM-VF-0013 marksL3 ExamCalculatorResistance
      A 22 kg particle moves with resistance 4v4v N only, starting at 1010 m s−1^{-1}. Find its speed after 0.50.5 s (3 s.f.).

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FM-VF-0023 marksL3 ExamCalculatorv dv/dx
      A 11 kg particle is repelled from OO by a force 8x2\frac8{x^2} N. It starts at rest at x=1x=1. Find its speed at x=4x=4 (3 s.f.).

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FM-VF-0033 marksL3 ExamNon-calculatorResistance
      A 22 kg particle moves with resistance 4v4v N only, starting at 1010 m s−1^{-1}. Find the total distance it travels.

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-FM-VF-0042 marksL4 ChallengingCalculatorResistance
      A 7070 kg skydiver falls with air resistance 0.25v20.25v^2 N. Find the terminal speed in m s−1^{-1} (g=9.8g=9.8, 3 s.f.).

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • a=dvdt=vdvdxa=\frac{dv}{dt}=v\frac{dv}{dx}.
      • Separate and integrate.
      • Resistance opposes motion.
      Section 12 of 14

      Quick quiz

      Q1 · SA-FM-VF-Q011 markL1 FoundationNon-calculatorv dv/dx
      If the force depends on displacement xx, write acceleration as
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      A level Further Mathematics 9FM0
      Where
      Further Mechanics 2 · Further dynamics
      Expected
      Newton’s second law with variable forces; a = dv/dt = v dv/dx; resisted motion; differential equations.
      Calculator
      Permitted; show enough working to earn the method marks.
      Typical style
      Edexcel often opens with a routine part and builds to a "show that" or "prove" part.
      Open in
      YMA01 · WME039FM0 · Option
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

      Browse past papers

      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel YMA01, 2018 specification (official specification).